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2 years ago

Would you rather cuddle a koala or pal around with a panda?

Physics

2 answers:

Marianna [84]2 years ago

6 0

Answer:

I don't really understand the wording of this question but I'd totally cuddle a Koala! If they weren't so mean

Explanation:

Thanks for listening!

vivado [14]2 years ago

4 0

Pal around with a panda. Anyone who says otherwise is severely biased in their judgement. Why would you cuddle with a koala that might not even like being around you when you can interact with another bear that will actually bond with you. Pandas are so much better because they eat bamboo. Name one interesting thing about a koala before you make your decision. Exactly. There are none. Pandas are superior to koalas and I would much rather pal around with a panda than cuddle with a koala.

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Refer to the first diagram. What is the weight of the person hanging on the end of the seesaw in Newtons?

irina1246 [14]

Due to equilibrium of moments:

1) The weight of the person hanging on the left is 250 N

2) The 400 N person is 3 m from the fulcrum

3) The weight of the board is 200 N

Explanation:

To solve the problem, we use the principle of equilibrium of moments.

In fact, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

The moment of a force is defined as:

$M=Fd$

where

F is the magnitude of the force

d is the perpendicular distance of the force from the fulcrum

In the first diagram:

- The clockwise moment is due to the person on the right is

$M_c = W_2 d_2$

where $W_2 = 500 N$ is the weight of the person and $d_2 = 2 m$ is its distance from the fulcrum

- The anticlockwise moment due to the person hanging on the left is

$M_a = W_1 d_1$

where $W_1$ is his weight and $d_1 = 4 m$ is the distance from the fulcrum

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the weight of the person on the left:

$W_1 d_1 = W_2 d_2\\W_1 = \frac{W_2 d_2}{d_1}=\frac{(500)(2)}{4}=250 N$

Again, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment due to the person on the right is

$M_c = W_2 d_2$

where $W_2 = 400 N$ is the weight of the person and $d_2$ is its distance from the fulcrum

- The anticlockwise moment due to the person on the left is

$M_a = W_1 d_1$

where $W_1 = 300 N$ is his weight and $d_1 = 4 m$ is the distance from the fulcrum.

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the distance of the person on the right:

$W_1 d_1 = W_2 d_2\\d_2 = \frac{W_1 d_1}{W_2}=\frac{(300)(4)}{400}=3 m$

As before, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment around the fulcrum this time is due to the weight of the seesaw:

$M_c = W_2 d_2$

where $W_2$ is the weight of the seesaw and $d_2 = 3 m$ is the distance of its centre of mass from the fulcrum

- The anticlockwise moment due to the person on the left is

$M_a = W_1 d_1$

where $W_1 = 600 N$ is his weight and $d_1 = 1 m$ is the distance from the fulcrum

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the weight of the seesaw:

$W_1 d_1 = W_2 d_2\\W_2 =\frac{W_1 d_1}{d_2}= \frac{(600)(1)}{3}=200 N$

#LearnwithBrainly

8 0

3 years ago

Really confused on this. Any help will be great

pav-90 [236]

C is what i would go with

8 0

3 years ago

What is the efficiency of a machine?

Julli [10]

Answer:

Efficiency is the percent of work put into a machine by the user (input work) that becomes work done by the machine (output work).

Explanation:

It is a measure of how well a machine reduces friction.

7 0

3 years ago

A charge alters the space around it. What is this alteration of space called?

iogann1982 [59]

Answer: electric field

Explanation: when a charge is placed in space, it alters the space around it by creating an electric field.

This electric field has the ability to exert a force (f) on any test charge(q) placed within this vicinity.

This is the reason why a charge can either attract or repel another charge.

6 0

3 years ago

what was the mass of a cannkn ball whose velocity is 200m/s if it were shot from 1000kg that recoils a 2m/s. Solve step by step

Whitepunk [10]

The famous Newton’s Third Law states that “For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.”

By using this,

10grams or 0.01kg of bullet with speed 400 m/sec and 5kg gun recoil with speed suppose ‘v’.

0.01×400=5×v

4/5=v

v=0.8m/sec ANSWER.

6 0

3 years ago